Apply Taylor's formula to f(x)=e^x expanded about a=0

Apply Taylor's formula to $f(x)=e^x$ expanded about $a=0$.

Step 1: Taylor's formula

With Lagrange remainder, Taylor's formula about $a$ reads $f(x)=\sum_{k=0}^n\dfrac{f^{(k)}(a)}{k!}(x-a)^k+R_n(x)$, $R_n(x)=\dfrac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1}$, $\xi$ between $a$ and $x$. Setting $a=0$ gives the Maclaurin series.

Step 2: Derivatives of $f(x)=e^x$

Since $\dfrac{d^k}{dx^k}e^x=e^x$ for every $k\ge0$: \[f^{(k)}(x)=e^x, \qquad f^{(k)}(0)=e^0=1 \quad\text{for all }k\ge0. \tag{1}\]

Step 3: Taylor polynomial

Substituting (1) with $a=0$: \[e^x = \sum_{k=0}^n\frac1{k!}x^k+R_n(x) = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^n}{n!}+R_n(x). \tag{2}\]

Step 4: Lagrange remainder

Since $f^{(n+1)}(x)=e^x$: \[R_n(x) = \frac{e^\xi}{(n+1)!}x^{n+1}, \qquad \xi\text{ strictly between }0\text{ and }x. \tag{3}\]

Step 5: Convergence

For any fixed $x$, $|e^\xi|\le e^{|x|}$ is a fixed constant, so

\[ |R_n(x)| \le \frac{e^{|x|}}{(n+1)!}|x|^{n+1} \to 0 \quad(n\to\infty), \tag{4} \] since $|x|^{n+1}/(n+1)!\to0$ for every finite $x$. The series converges to $e^x$ for all $x\in\mathbb R$.

Step 6: Verification

\[x=1,\ n=4\text{: }P_4(1)=1+1+\tfrac12+\tfrac16+\tfrac1{24}=2.708\overline3,\ e\approx2.71828,\ |{\rm error}|\approx0.0100\ \checkmark\]

\[x=1,\ n=9\text{: }P_9(1)\approx2.718281526,\ |{\rm error}|\approx3.0\times10^{-7}\text{, shrinking exactly as the bound }e/(n+1)!\text{ predicts}\ \checkmark\]